Question: $f(t) = 2t^{2}+3t+5+4(h(t))$ $h(x) = x^{2}+6x+2$ $ h(f(0)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 2(0^{2})+(3)(0)+5+4(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = 0^{2}+(6)(0)+2$ $h(0) = 2$ That means $f(0) = 2(0^{2})+(3)(0)+5+(4)(2)$ $f(0) = 13$ Now we know that $f(0) = 13$ . Let's solve for $h(f(0))$ , which is $h(13)$ $h(13) = 13^{2}+(6)(13)+2$ $h(13) = 249$